1. We can use the formula a^2=b^2+c^2-2bcCos(A) for this question. To find out the longest length possible for the 3rd side we want angle A to be as large as possible, so A will be 179.9recurring, if we then assign b to 6, c to 10, and a to the third side, we can calculate that the longest possible length of a will be 15.9recurring from 15.9^2=6^2+10^2-2x6x10xCos(179.9recurring). As the question asks for the largest possible integer we have to round 15.9recurring down to 15, therefore the answer is 15. I'm not too sure about this one though as this is either a really abstract question or i am missing something big out.
2. It's kind of hard to explain this one without a visual representation but basically due to the rules of bisectors BC/CD=BA/AD, BC/12=18/8, BC=12x18/8, BC=27
3. The perimeter of the triangle is equal to the sum of the length of all 3 sides, in this case due to the 1:2:3 ratio we can say one side has length l, one side has length 3l, and the third side has length 5l, therefore perimeter equals 5l, 5l=28, l=5.6, 5.6 is the length of the shortest side.
